¨ ¨ ¨ HF´pS3q À
j”i mod pHF´pS30pT q, jq HF´pS3ppT q, iq ¨ ¨ ¨
¨ ¨ ¨ HF8pS3q À
j”i mod pHF8pS30pT q, jq HF8pS3ppT q, iq ¨ ¨ ¨
¨ ¨ ¨ HF`pS3q À
j”i mod pHF`pS30pT q, jq HF`pS3ppT q, iq ¨ ¨ ¨
... ... ...
Using these exact grids, we can, for example, compute HF`pΣp2, 3, 5qq “ HF`pP q.
It turns out that HF`pP q “ ZrU´1s, with dpP q “ 2. (This is detailed in Ozsv´ath and Szab´o’s Absolutely graded Floer homologies and intersection forms for four-manifolds with boundary.) Recall that we used this fact earlier in our proof that Θ3Z has a Z summand!
We’ve just shown that d{2 : Θ3
ZÑ Z is surjective, since it sends P to 1. We can also show that HFredpP q “ 0.
6.3 General algorithm to compute y HF (lecture 12)
As mentioned a few times prior, one of the most powerful aspects of Heegaard Floer ho-mology is that it can theoretically be computed for any space, if we know its Heegaard decomposition! This means that any question answered in terms of Heegaard Floer ho-mology answers the question in general. (For example, we will see in the next lecture that Heegaard Floer homology detects the Thurston norm, i.e. lower bounds on the genus of surfaces representing a class in H2pY3q.)
Theorem 6.3.1 (Sarkar-Wang). Let H be a Heegaard diagram of a 3-manifold Y . There is an algorithm to compute yHF pY q in Z{2Z-coefficients in terms of the combinatorial data of H.
This theorem will take a lot of work to prove, so we’ll start by explaining the goal.
That is, what exactly do we need to create an algorithm for? The problem comes down to computing the differentials in the Heegaard Floer chain complex. This map cares about
ϕ P π2px, yq, µpϕq “ 1 and Dpϕq “ÿ
i
aiRi, ai ě 0.
The first condition on ϕ is immediate from the definition of the map, but the second condition comes from implicit complex geometry: one can show that
ai “ rϕs ˝ rtziu ˆ Symg´1pΣqs, zi P Ri,
and for an arbitrary ϕ P π2px, yq this can be anything. However, for ϕ to have a holomorphic representative, ai must be non-negative.
To understand the boundary maps, we must calculate #pMpϕq{Rq for ϕ as above. We have discovered earlier that this is difficult in general! The idea is to change the Heegaard diagram (by Heegaard moves) to make it easy to count. Recall from earlier that we know that if a domain Dpϕq is a rectangle or bigon, then #pMpϕq{Rq “ ˘1. If we work in mod 2 coordinates, all signs are lost! Therefore we will aim to arrange Heegaard diagrams in this manner - then the boundary map can genuinely be computed combinatorially. The main idea of the theorem is that all Heegaard diagrams can be changed into a Heegaared diagram in which all regions (not including the basepoint) are either bigons or rectangles!
Expanding and proving all of these ideas will be the goal of the rest of this section.
Remark. The result from this section is actually very surprising. Lagrangian Floer ho-mology is known to require geometry and gauge theory, as well as analysis. Since yHF was modelled off Lagrangian Floer homology, it was also expected to depend on these tools.
This result started a whole new direction of research.
The remainder of this section can be broken up into two subsections:
1. Proof that #pMpϕq{Rq “ ˘1 for Dpϕq consisting of bigons or rectangles, and µpϕq “ 1.
2. Proof that Heegaard diagrams are equivalent to ones with just bigons and rectangles (away from the basepoint).
As mentioned above, combining these two facts leads to the desired result.
Lemma 6.3.2. Let Dpϕq “ř
iaiRi be a domain, with ai ě 0 for each i, and Ri a bigon or rectangle whenever ai ą 0. Suppose moreover that µpϕq “ 1. Then Dpϕq has exactly one J -holomorphic representative.
Proof. Let D be a domain for ϕ P π2px, yq as above. Recall the Lipshitz formula:
µpϕq “ epDq ` nxpDq ` nypDq,
where epDq is the Euler measure, and nx and ny are average vertex multiplicities. Since x, y P TαX Tβ, we can write x “ tx1, . . . , xgu and y “ ty1, . . . , ygu, where the xi and yj are genuine intersection points of curves on Σ.
Recall that domains D satisfy BD “ BαD ` BβD, where BpBαDq “ BpBβDq “ x ´ y “ÿ
xi´ ÿyi.
We now calculate some relations that must hold between the multiplies of regions meeting at a vertex. Suppose p P αi X βi is an intersection point, with xi R x, p R y.
Locally D looks like the left side of figure 6.3: notice that since p is not a vertex of D, any multiplicity at a quadrant must be shared with neighbouring quadrants (so that no
“corners” occur). Overall, the region near the quadrant must be formed by laying “strips”
adjacent to p.
Figure 6.3: Local structure at p, where p P αjX βi does not lie in x nor y.
Using the coordinates on the right, we have
a “ q ` r, b “ p ` r, c “ q ` s, d “ p ` s.
But now a ` d “ p ` q ` r ` s “ b ` c. Using similar analyses together with the local examples in figure 6.4 (to show the choice of orientation), we obtain the following identities:
p R x, p R y ñ a ` d “ b ` c
p R x, p P y ñ a ` d “ b ` c ` 1
p P x, p R y ñ a ` d ` 1 “ b ` c
p P x, p P y ñ a ` d ` 1 “ b ` c ` 1 ñ a ` d “ b ` c.
In particular, in the first and last case, the average vertex multiplicity at p is a half-integer, while in the middle cases it is an odd quarter-integer (i.e. pa ` b ` c ` dq{4 P 12Z `14).
We can now proceed with the main body of the proof. Let Dpϕq “ ř
iaiRi with ai ě 0, and Ri a bigon or rectangle whenever ai ą 0. Suppose µpϕq “ 1. Two possible types of domains satisfying these conditions are shown in figure 6.5. We know from earlier in the class that domains of the types shown in figure 6.5 have exactly one J -holomorphic representative. Therefore we prove that no other types of domains are possible!
By the Lipshitz formula, µpϕq “ epDq ` nxpDq ` nypDq “ 1. Since epDq “ř
iaiepRiq, the only Euler measure contributions come from regions with ai ą 0. Such regions are
Figure 6.4: Local structure at xi and yi (example).
Figure 6.5: Two types of domains with one holomorphic representative.
necessarily bigons or rectangles, which have Euler measure 0 or 1/2. Recall that this comes from the formula
epΣq “ χpΣq ´ #right angles
4 .
We also have the assumption that nxpDq, nypDq ě 0, from the fact that each ai is non-negative. This immediately restricts us to three cases:
1. epDq “ 1, nxpDq “ nypDq “ 0.
2. epDq “ 1{2, nxpDq ` nypDq “ 1{2.
3. epDq “ 0, nxpDq ` nypDq “ 1.
We will now work through these cases one by one.
1. If nxpDq “ nypDq “ 0, then there are no points p P αi X βj X D such that p lies in exactly one of x or y. This is because of the local structure of intersection points determined at the start of this proof: the average vertex multiplicity of any such point is necessarily an odd quarter integer (which is necessarily non-zero). The sum of any number
of non-negative non-zero numbers will give a positive result, which is prohibited. Therefore all intersection points p occuring in D either belong to neither x nor y, or both x and y.
This forces x “ y for our strip ϕ, so that ϕ is the constant strip. Then µpϕq “ 0. This breaks one of our premises! No domains D are of the first type.
2. If epDq “ 1{2 and nxpDq ` nypDq “ 1{2, then either nxpDq “ nypDq “ 1{4, or one of them is 0 and the other is a half. We also know that the regions making up the domain must consist of exactly one bigon and many rectangles, simply by considering epDq. The existence of the bigon prohibits the latter options for nxpDq and nypDq, so we must have nxpDq “ nypDq “ 1{4. Now there are exactly two intersection points which are non-constant, and the rest have trivial multiplicity. Overall this tells us that D must itself be a bigon, i.e. of the form described earlier.
3. In the final case, epDq “ 0 and nxpDq ` nypDq “ 1. Since epDq “ 0, all regions are necessarily rectangles. The only way rectangles can be stitched together to achieve the premise is for D itself to be a large rectangle, i.e. of the form described earlier.
This completes the proof that our domain Dpϕq is of a form where #pMpϕq{Rq is known to be plus or minus 1!
The next question is whether or not we can take an arbitrary Heegaard diagram and turn it into a diagram where all regions are bigons or rectangles. If so, that would be our algorithm!
Remark. Unfortunately, this is impossible in general. Suppose the Heegaard diagram has genus gpΣq ě 2. Then
2 ´ 2g “ χpΣq “ epΣq “ÿ
epRiq ě 0,
which is absurd. Moreover, the only manifolds with Heegaard diagrams of genus at most 1 are S3, S1ˆ S2, and lens spaces Lpp, qq.
Remark. Rejoice! For yHF we only require counting regions that do not include the basepoint z P Σ! This means we cna hopefully transform our Heegaard diagram into one for which all regions not including the basepoint are either bigons or squares. This can indeed be done.
Theorem 6.3.3. Let Y have based Heegaard diagram pΣ, α, β, zq. Then one can change α, β by a sequence of handleslides and isotopies to obtain a Heegaard diagram in which all regions not containing z are bigons or rectangles. (Sarkar and Wang refer to such diagrams as nice diagrams).
Proof. Let pΣ, α, β, zq be the Heegaard diagram. Originally, some regions might not be polygons (that is, they might not be simply connected). By using isotopies, we can assume without loss of generality that all regions are polygons. (See figure 6.6.)
Figure 6.6: Using isotopies, all regions can be made into polygons.
Each region R is now a polygon, with 2n edges. We say that R is good if n ď 2, and bad if n ě 3. This is because we want our regions to eventually be bigons or rectangles, which are exactly the good polygons. We define the badness of a polygon to be
bpRq “ maxp0, n ´ 2q.
Thus the goal of the proof is to move all of the badness to the region containing z.
Now let D denote a bad region. We further define dpDq to be the minimum number of β curves that must be crossed to get from D to the region containing z, by a path not intersecting α-curves. Intuitively, dpDq is the distance from D to z through β curves.
Figure 6.7: Using isotopies, all regions can be made into polygons.
In the left of figure 6.7, the region D has distance dpDq “ 4. (It also has badness bpDq “ 1). The main idea of this proof is not to initially decrease the badness itself, but decreases the distances of the bad regions. For example, on the right of figure 6.7, we have
“pushed a finger” from a β curve through a series of α curves. In the resulting Heegaard
diagram, there are two bad regions instead of one; of badnesses 1 each. However, the bad regions have distance 1 and 3 from z, so maxtdpDq : bpDq ‰ 0u has decreased!
dpD0q “ 4, bpD0q “ 1 ù
#
dpD1q “ 3, bpD1q “ 1 dpD2q “ 1, bpD2q “ 1.
In our algorithm, we will aim to move all badness closer to z, even at the expense of the number of bad regions or individual badnesses getting worse.
More precisely, let d “ maxtdpDq : bpDq ‰ 0u denote the maximum distance of any bad region from z. Let D1, . . . , Dm be the bad regions at distance d from z. We want a sequence of moves doing the following:
1. Decrease the maximum distance d.
2. Keep the same d, but decrease the total badness of the regions at distance d, i.e.
decreaseřm
i“1bpDiq.
3. Keep the same d andřm
i“1bpDiq, but decrease m, i.e. decrease the number of regions achieving the maximum distance d. Formally we will decrease the distance complex-ity, which is the quantity pbpD1q, . . . , bpDmqq measured with the lexicographic norm, where Diare initially ordered so that bpDiq ď bpDi`1q. Notice that if this strictly de-creases with each step, then after a finite number of steps m can be made arbitrarily small.
We order these changes lexicographically - each is an improvement to the diagram, but decreasing d is preferred over decreasing řm
i“1bpDiq etc. Therefore if we consider the quantity
with lexicographic norm, our goal is to find a sequence of moves which continuously de-creases pd,řm
i“1bpDiq, bpD1q, . . . , bpDmqq until d “ 0. We will refer to the lexicographic size of pd,řm
i“1bpDiq, bpD1q, . . . , bpDmqq as the complexity of the diagram.
We will now describe a collection of such moves (which forms the algorithm), hence completing the proof.
1. Choose a bad region with the least badness amongst regions achieving the maximum distance d. (That is, consider the region D1.)
2. Push a finger from a β curve bounding the region through an α curve of the same region, and continuing pushing as follows:
(a) If the finger enters a rectangle, keep pushing - unless the rectangle has a smaller distance to z.
(b) If the finger enters a bigon, stop.
(c) If the finger enters a bad region, stop.
(d) If the finger re-enters the original region, then it necessarily encircles a β curve.
Handleslide over the encircled β curve, as in figure 6.8.
Figure 6.8: Handleslide, if a finger is pushed into its starting region.
The moves above are guaranteed to strictly decrease the completely of the Heegaard diagram with every move! Therefore it must eventually force d “ 0, as required. It remains to verify that the moves really decrease the complexity.
If we stop at (a), then the original region has been broken into two regions, with smaller badness. The complexity has strictly decreased because d could not have increased, since the only region with increasing badness has distance less than d from z. However, if d has not decreased, thenř bpDiq has strictly decreased, as the original region has been broken into separate regions! (One can verify that it has decreased by exactly 1.)
If we stop at (b), the same argument as above holds.
If we stop at (c), by the hypothesis, the region in which we’ve stopped has a smaller-or-equal distance to z. If the region has a smaller distance, the same argument as above holds. If the region has the same distance, then d has not decreased, and one can verify thatř bpDiq is also left unchanged. However, we have moved some of the badness of D1 to a subsequent region, so the complexity has indeed strictly decreased.
Finally if we stop at (d), then once against d might have decreased, but if not, then ř bpDiq has decreased!
This completes the proof! (Technically some of the details are missing in the difficult case of (d), but these can be checked in the original paper of Sarkar and Wang.)
Corollary 6.3.4. There is a combinatorial description of yHF over Z{2Z.
Later we will study some other algorithms as well! In particular, Manolescu, Ozsv´ath and Dylan Thurston developed an algorithm for HF˘ and 4-manifold invariants, using surgery formulae and nice diagrams.
Chapter 7
Applications in three dimensions
In an earlier chapter we studied some applications of Heegaard Floer homology in four dimensions, but we have yet to see any three dimensional applications! One of the most celebrated applications is that it detects the Thurston norm. We will also study L-spaces, which are the “homologically simple spaces” in the setting of Heegaard Floer homology.
7.1 L-spaces (lecture 13)
Definition 7.1.1. A 3-manifold Y is an L-space if yHF pY, sq “ Z for all s in SpincpY q.
This is the Heegaard Floer homology version of an integral homology sphere.
Example. Some L-spaces include S3, lens spaces, and the Poincar´e homology sphere.
Example. Some spaces which are not L-spaces are 1{n-surgeries on the trefoild knot T , for n ě 2. One can also simply consider spaces with b1 ą 0.
Proposition 7.1.2. Let Y be a closed oriented 3-manifold. The following are equivalent:
1. Y is an L-space.
2. HF`pY, sq “ ZrU´1s for all s.
3. HF´pY, sq “ ZrU s for all s.
4. Y is a rational homology sphere, and HFrefpY q “ 0.
Corollary 7.1.3. If a closed 4-manifold X admits an L-space as an admissible cut, then ΦX,s“ 0 for all s.
Proof. We now prove the previous proposition. The first three are all equivalent via the exact triangles
0 Ñ yHF Ñ HF˘ UÝÑ HF˘ Ñ 0.
It remains to prove that Y is an L-space if and only if Y is a rational homology sphere with trivial reduced Heegaard Floer homology.
One direction is easy: suppose Y is a rational homology sphere. Then HF8pY, sq “ ZrU, U´1s,
from which we have that HF´pY, sq “ ZrU s‘HFred. Now if HFredpY q vanishes, HF´pY, sq “ ZrU s as required.
For the converse, we require a lemma. Namely, the Euler characteristic χp yHF pY, sqq depends only on homology: so Y is a rational homology sphere.
We now prove this lemma. Recall that HF pY q “y à
sPSpincpY q
HF pY, sq,y and that
χp yHF pY qq “ rTαs ¨ rTβs in SymgpΣq.
A Heegaard decomposition of genus g for Y consists of a CW-complex for Y with one 0-cell, g 1-cells, g 2-cells, and one 3-cell. Therefore H˚pY q comes from the complex
0 Ñ ZÝÑ Z0 g DÝÑ Zg 0ÝÑ Z Ñ 0,
The remaining ingredient is the fact that χp yHF pY, sqq is independent of s! This is because we can change s by moving the basepoint to a different location. (That is, the bijective
correspondence of SpincpY q with SpY q is non-canonical, depending on the choice of base
This motivates L-spaces in one way: understanding L-spaces helps us to understand 4-manifolds with trivial mixed invariant.
Another reason we care about L-spaces is because of the Boyer-Gordon-Watson conjec-ture, which we first state and then explain (since the statement will contain some unfamiliar terms).
Conjecture. Let Y be a closed oriented 3-manifold. It is conjectured that the following are equivalent:
1. Y is an L-space.
2. Y carries a coorientable taut foliation.
3. π1pY q is left orderable.
Remark. This conjecture, if it holds, is truly remarkable. The first statement lies in the realm of Heegaard Floer homology, which comes from symplectic geometry (and has a combinatorial description in terms of a Heegaard decomposition). The second statement lies in the realm of differential topology, and the third statement is purely algebraic. The conjecture binds together separate mathematical realms.
We now describe what the conjecture actually means, by defining the missing termi-nology.
Definition 7.1.4. A foliation of an n-manifold X is a decomposition X “Ť
tPTLt into k-dimensional leaves Lt, so that locally T satisfies
Rn“ ď
tPRn´k
pttu ˆ Rkq.
A foliation of a 3-manifold Y3 by surfaces is said to be taut if there is an embedded circle γ P Y meeting every leaf transversely at least once. Finally, any foliation is said to be coorientable if the normal bundles to the leaves are orientable.
Example. The product S1ˆ Σ for Σ a surface is trivially a foliation by surfacesttu ˆ Σ.
It’s automatically taut, as we can take the curve S1ˆ tptu for any point in Σ.
Example. A non-example of a taut foliation is the Reeb foliation, which is a foliation of S3 by surfaces by decomposing it into two solid tori, and foliating each solid torus by spiralling “cups”.
There are some cool results about taut foliations, such as the following:
Theorem 7.1.5 (Sullivan). A foliation of Y3 by surfaces is taut if and only if there exists a metric on Y such that all leaves are minimal surfaces.
Theorem 7.1.6 (Eliashberg-Thurston). If a manifold Y3 admits a taut foliation, then it admits a tight contact structure.
We do not explain what the above theorem means, but Ozsv´ath and Szab´o proved that if a space admits a tight contact structure then it cannot be an L-space! This is the only direction known to hold in the Boyer-Gordon-Watson conjecture.
Next we briefly explore the algebraic side of the conjecture.
Definition 7.1.7. A non-trivial group G is called left-orderable if there exists a total order on G such that g ă h if and only if kg ă kh, for all k P G.
Example. Z, R, and Zn (lexicographical) are all left orderable.
Example. A non-example is any group G with non-trivial torsion. Suppose g P G has order n. Then if 1 ă g, then g ă g2, and g2 ă g3 and so on, until we reach gn´1 ă gn“ 1.
Therefore 1 ă 1, which is impossible. The same contradiction holds if g ă 1.
The BGW conjecture has actually been verified for many classes of 3-manifolds! For example, all graph manifolds are known to satify the conjecture. These are manifolds that split into Seifert fibred spaces after cutting along spheres and tori.
Finally we remark that L-spaces are rational homology spheres with trivial reduced Heegaard Floer homology, so there is no clear connection with integral homology spheres.
It turns out that the only known integral homology spheres which are L-spaces are obtained by gluing together the Poincar´e homology sphere.
Conjecture. If Y is a ZHS3, then Y is an L-space if and only if Y is a connected sum
Conjecture. If Y is a ZHS3, then Y is an L-space if and only if Y is a connected sum